Question

What is 7 over (fraction bar) (8×4)+90÷9 ÷ (6^3)^-2 × 9 over (fraction bar) 6(5+4) = ?????

Answer 1

The answer is: " 36 " .
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Note: " (8 * 4) + 90 ÷ 9 " = " (8 * 4) + (90 ÷ 9) " = "32 + 10 = 42" .

We can rewrite the "fraction":

7 / [ (8 * 4) + 90 ÷ 9 ] ; as:
`(7)/(42)` ;

and can reduce this value to (and rewrite as): "
`(1)/(6)` " ;
{ since: "
`(7)/(42)` =
`(7/7)/(42/7)` =
`(1)/(6)`" .}.

Note: The "fraction" :
`(9)/(6(5+4))` ; the "9/6" cancel to "3/2" ;
{Since: "9÷3 = 3" ; and since: "6÷3 = 2" ).

So rewrite:
`(9)/(6(5+4))` ; as:


`(3)/(2(5+4))` ;

Note:
`(3)/(2(5+4))` ;

=
`(3)/(2(9))` ;

=
`(3)/(18)` ;

→ Note:
`(3)/(18)` ; can be reduced to; & rewritten as:

"
`(1)/(6)` " ;

→ {since: "
`(3)/(18)` =
`(3/3)/(18/3)` ;

=
`(1)/(6)` ;
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Note that: (6²) ⁻² = 6⁽ ² * ⁻²⁾ = 6⁻ ⁴ ;

=
`(1)/(6^4)` ;
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So, we can rewrite our expression as:


`(1)/(6)` ÷
`(1)/(6^4)` *
`(1)/(6)` ;
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Note:
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`(1)/(6)` ÷
`(1)/(6^4)` *
`(1)/(6)` ;

= (
`(1)/(6)` ÷
`(1)/(6^4)` ) *
`(1)/(6)` ;

= (
`(1)/(6)` *
`(6^4)/(1)` ) *
`(1)/(6)` ;

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Start with :
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→ (
`(1)/(6)` *
`(6^4)/(1)` ) ;

The "6" in the "
`(1)/(6)` " ; cancels to a "1" ;
and the "6⁴ " in the "
`(6^4)/(1)` ) " ; cancels to a "6³ " ;

→ {Since: "6⁴ / 6 = 6⁴ / 6¹ = 6⁽⁴ ⁻ ¹⁾ = 6³ ;

→ and we have:
`(1)/(1)` *
`(6^3)/(1)` ;

→ The "
`(1)/(1)` " can be eliminated; since "1÷1= 1 " ;
and since: any value, divided by "1", equal that same value; & since "any value, multiplied by "1"; equals that same value" ;
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We have: " 6³ " .
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Now, we multiplied this value, " 6³ " , by "
`(1)/(6)` " ;

→ 6³ *
`(1)/(6)` ;

=
`(6^3)/(1)` *
`(1)/(6)` ;

Note: The "6³ " cancels out to "6² " ; and the "6" cancels out to "1" ;

{since: "6³ ÷ 6 = 6³ ÷ 6¹ = 6⁽³ ⁻ ¹⁾ = 6² " ;
and since: "6 ÷ 6 = 1 " ; ________________________________________________________
→ The answer is: 6² = 6 * 6 = 36 ;
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The answer is: " 36 " .
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